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Dermal Exposure Under Occlusion and Wash-Off

Learn how to represent occlusion, application-area changes, and wash-off in a dermal surface–reservoir model, while checking absorption timing against mass balance.

Intermediate 12 min read

What You'll Learn

  • Write dermal surface and reservoir equations for constant exposure conditions.
  • Calculate cumulative and instantaneous absorption under different occlusion assumptions.
  • Relate application area to the total permeation rate and distinguish fixed total dose from fixed dose per area.
  • Model an instantaneous partial wash-off and track the removed, retained, and absorbed amounts.
  • Use mass balance and limiting cases to diagnose implausible dermal input calculations.

Why Exposure Conditions Change Dermal Input

A dermal dose is not necessarily absorbed at the same rate under every exposure condition. Occlusion can change hydration, temperature, and contact between formulation and skin; changing application area can change the total permeating amount per unit time; and washing the surface can remove material that has not yet entered the skin. These changes affect the input to the body, not merely the amount written on the dose label.

The model here extends a surface-dose and skin-reservoir description. It keeps surface material distinct from material already in the skin, because washing can remove the former but does not instantaneously remove the latter. Its rate constants are effective parameters for a specified formulation, area, and exposure condition. Occlusion does not imply one universal direction or magnitude of change: the relevant parameters must be measured or justified for the application being modelled.

Definition: Let \(S(t)\) be the drug amount remaining on the skin surface, \(R(t)\) the amount in a skin reservoir available for onward absorption, and \(A(t)\) the cumulative amount absorbed into the systemic circulation, all in mg. Let \(k_p\) be the first-order transfer rate constant from surface to reservoir and \(k_r\) the first-order rate constant from reservoir to systemic circulation, both in h\(^{-1}\). The instantaneous systemic input rate is \(k_rR(t)\), in mg/h.

Baseline Surface–Reservoir Model

For a single application with initial surface amount \(S_0\), assume \(S(0)=S_0\), \(R(0)=0\), and \(A(0)=0\). Between interventions, assume constant \(k_p>0\) and \(k_r>0\), first-order transfer at each step, no surface loss, no degradation, and no other route of removal. These are simplifying assumptions: they make the amount leaving the surface enter the reservoir, and the amount leaving the reservoir enter the systemic circulation.

$$ \frac{dS}{dt}=-k_pS,\qquad \frac{dR}{dt}=k_pS-k_rR,\qquad \frac{dA}{dt}=k_rR. $$

Solving the first equation gives \(S(t)=S_0e^{-k_pt}\). Substitution into the reservoir equation gives a linear differential equation. Multiplying it by the integrating factor \(e^{k_rt}\), integrating from 0 to \(t\), and using \(R(0)=0\) yields the following result when \(k_p\ne k_r\).

$$ R(t)=S_0\frac{k_p}{k_r-k_p} \left(e^{-k_pt}-e^{-k_rt}\right),\qquad A(t)=S_0-S(t)-R(t). $$

The second equation follows from mass balance: under these assumptions, \(d(S+R+A)/dt=0\), so the sum remains \(S_0\). If \(k_p=k_r=k\), the equal-rate limit is \(R(t)=S_0kt e^{-kt}\); the expression with \(k_r-k_p\) in its denominator must not be evaluated by direct substitution in that case.

Key distinction: \(R(t)\) is an amount in the skin, not an amount already absorbed. The systemic input rate is \(k_rR(t)\), whereas the cumulative absorbed amount is \(A(t)\). A change in the rate constants can change the timing of input even when the eventual absorbed amount is unchanged.

Occlusion as a Change in Exposure Parameters

In this model, an occluded condition is represented by using parameters appropriate to occlusion, such as \(k_{p,\mathrm{occ}}\) and \(k_{r,\mathrm{occ}}\), in place of the unoccluded values. Occlusion may alter permeation, reservoir release, or both. It is not valid to assume from the word “occluded” alone that either rate constant must increase. The equations are applied separately under each condition, with its own parameter values and the same initial dose if the comparison holds dose constant.

Worked Example: Comparing Occluded and Unoccluded Input

Suppose a hypothetical 100 mg surface dose is applied at \(t=0\). For illustration, let \(k_r=0.20\) h\(^{-1}\) in both conditions, with \(k_p=0.40\) h\(^{-1}\) without occlusion and \(k_p=0.60\) h\(^{-1}\) under occlusion. Assume the baseline model’s no-loss conditions hold. Find the amounts in the surface and reservoir, cumulative absorption, and systemic input rate at 2 h.

Without occlusion, \(S(2)=100e^{-0.40(2)}=44.93\) mg. The reservoir amount is

$$ R(2)=100\frac{0.40}{0.20-0.40} \left(e^{-0.80}-e^{-0.40}\right) =44.20\ \text{mg}. $$

Thus \(A(2)=100-44.93-44.20=10.87\) mg, and the input rate is \(0.20(44.20)=8.840\) mg/h.

Under occlusion, use that condition’s own \(k_p=0.60\) h\(^{-1}\) throughout the formula. Then \(S(2)=100e^{-1.20}=30.12\) mg, and

$$ R(2)=100\frac{0.60}{0.20-0.60} \left(e^{-1.20}-e^{-0.40}\right) =55.37\ \text{mg}. $$

Therefore \(A(2)=100-30.12-55.37=14.51\) mg, and the input rate is \(0.20(55.37)=11.07\) mg/h. In this hypothetical parameterization, occlusion is associated with 3.643 mg more absorbed by 2 h and a higher input rate at 2 h. Those results arise from the chosen rate constants; they are not a general prediction that occlusion always increases absorption.

Application Area and the Meaning of Dose

Area changes require clarity about what is held constant. If the local surface concentration is \(C_s(t)\) and the permeation flux per unit area is \(J(t)\), a simple flux representation gives a total transfer rate of \(A_{\mathrm{app}}J(t)\), where \(A_{\mathrm{app}}\) is application area. In a first-order surface model, an effective rate constant may be written \(k_p=P A_{\mathrm{app}}/V_s\), where \(P\) is an effective permeability and \(V_s\) is an effective surface donor volume. This representation assumes those quantities are meaningful and approximately constant over the interval considered.

If the total applied amount is held fixed, increasing area does not increase the initial total amount, but it may change \(k_p\) and therefore timing. If the dose per unit area is held fixed, increasing area also increases total applied amount. These are different comparisons and should not be conflated. In real applications, area can also alter local depletion, formulation thickness, contact, and the applicability of a single well-mixed surface compartment.

Worked Example: Doubling Area at Fixed Total Dose

Consider a hypothetical 50 mg total surface dose, the same effective donor volume \(V_s=5\) cm\(^3\), permeability \(P=0.020\) cm/h, and reservoir release rate \(k_r=0.50\) h\(^{-1}\). Compare areas of 50 cm\(^2\) and 100 cm\(^2\) at 2 h. Assume the same total dose and the baseline no-loss model in both cases. From \(k_p=P A_{\mathrm{app}}/V_s\), the two surface transfer rates are \(0.20\) h\(^{-1}\) and \(0.40\) h\(^{-1}\), respectively.

For 50 cm\(^2\), \(S(2)=50e^{-0.40}=33.52\) mg and \(R(2)=50(0.20)/(0.50-0.20)(e^{-0.40}-e^{-1.00})=10.08\) mg. Hence \(A(2)=50-33.52-10.08=6.403\) mg.

For 100 cm\(^2\), \(S(2)=50e^{-0.80}=22.47\) mg and \(R(2)=50(0.40)/(0.50-0.40)(e^{-0.80}-e^{-1.00})=16.29\) mg. Hence \(A(2)=50-22.47-16.29=11.24\) mg. In this assumed model, spreading the same total dose over twice the area increases absorption by 2 h because the effective \(k_p\) doubled. This is a model-based illustration, not a universal area-scaling law.

Partial Wash-Off as an Intervention

Represent a wash at time \(t_w\) as an instantaneous removal of a fraction \(f\) of the surface amount, where \(0\le f\le1\). Immediately before washing, write the amounts as \(S(t_w^-)\) and \(R(t_w^-)\). Immediately afterward,

$$ S(t_w^+)=(1-f)S(t_w^-),\qquad R(t_w^+)=R(t_w^-). $$

The amount removed is \(fS(t_w^-)\). The skin reservoir is unchanged by this idealized surface wash; subsequent absorption continues from the retained surface amount and the pre-existing reservoir. After the wash, solve the same equations from the new starting amounts. This event-based treatment assumes the wash itself is instantaneous and does not change permeability or reservoir release. If washing changes those properties, the post-wash rate constants must also be changed.

Worked Example: Half of the Surface Dose Is Washed Off

Use \(S_0=100\) mg, \(k_p=0.40\) h\(^{-1}\), and \(k_r=0.20\) h\(^{-1}\), with no loss before the wash. At \(t_w=1\) h, wash away half the surface amount, so \(f=0.50\). Find the absorbed amount by 3 h and check the mass balance.

Just before washing, \(S(1^-)=100e^{-0.40}=67.03\) mg. The reservoir amount is \(R(1^-)=100(0.40)/(0.20-0.40)(e^{-0.40}-e^{-0.20})=29.68\) mg. Cumulative absorption before washing is \(A(1^-)=100-67.03-29.68\approx3.29\) mg (3.286 mg using unrounded values). The wash removes \(0.50(67.03)=33.52\) mg and leaves \(S(1^+)=33.52\) mg; \(R(1^+)=29.68\) mg.

Two hours later, at \(t=3\) h, the retained surface amount is \(S(3)=33.52e^{-0.40(2)}=15.06\) mg. The reservoir amount consists of the surviving pre-wash reservoir plus transfer from the retained surface:

$$ R(3)=29.68e^{-0.20(2)} +33.52\frac{0.40}{0.20-0.40} \left(e^{-0.40(2)}-e^{-0.20(2)}\right) =34.71\ \text{mg}. $$

By 3 h, \(A(3)=100-33.52-15.06-34.71=16.71\) mg, where the 33.52 mg subtracted is the amount washed away. Equivalently, 3.286 mg was absorbed before washing and \(33.52+29.68-15.06-34.71=13.43\) mg after washing, giving 16.71 mg total. The balance is \(33.52\) mg washed off, \(15.06\) mg on the surface, \(34.71\) mg in the reservoir, and \(16.71\) mg absorbed, which sum to \(100.00\) mg.

Derivation, Diagnostics, and Model Choice

The key justification for treating these changes separately is conservation of amount. Between interventions, transfer from surface to reservoir subtracts \(k_pS\) from one compartment and adds exactly \(k_pS\) to the other. Transfer from reservoir to systemic circulation subtracts \(k_rR\) and adds exactly \(k_rR\) to cumulative absorption. Thus \(d(S+R+A)/dt=0\). At an instantaneous wash, the amount removed, \(fS(t_w^-)\), must be recorded as a loss from the system; it cannot also be counted as absorbed.

1
Specify the comparison.
State whether total dose or dose per area is fixed, and identify the exposure condition and wash time.
2
Assign condition-specific parameters.
Use appropriate values of \(k_p\) and \(k_r\); if area is used to derive \(k_p\), state the assumptions behind that relationship.
3
Apply events to the correct compartment.
At a surface wash, reduce \(S\) by the stated fraction, leave \(R\) unchanged under the idealized assumption, and record the removed amount.
4
Check amounts and timing.
Calculate \(A(t)\) from the input rate or mass balance, and verify that nonnegative amounts plus absorption and recorded losses account for the applied dose.

A frequent diagnostic is a negative reservoir amount or a reservoir amount greater than the available dose. Either suggests a sign error, mismatched rate constants in the formula, or an inappropriate model assumption. In particular, each condition’s reservoir formula uses its own denominator \(k_r-k_p\). For example, pairing the \(e^{-0.60t}\) surface term with a denominator for \(k_p=0.40\) h\(^{-1}\) does not solve the stated equations and can create impossible amounts. Also check limiting behavior: with positive \(k_p\) and \(k_r\), no losses, and sufficient time, \(S\) and \(R\) approach zero and \(A\) approaches \(S_0\). If either transfer rate is zero, that conclusion need not hold.

This compartment model is a practical alternative to using a constant systemic input rate or treating the whole applied dose as immediately absorbed. Those simpler approaches may be adequate when only an average exposure is needed and absorption is rapid relative to the time scale of interest, but they do not represent surface depletion, reservoir delay, or wash-off explicitly. A spatial skin model can represent gradients and multiple layers more realistically, but needs more parameters and data. The two-compartment model is limited by its first-order and well-mixed assumptions; it does not by itself describe nonlinear permeability, spatially nonuniform application, changing formulation, or uncertainty in exposure conditions.

Common pitfall: It is wrong to say that a wash removes a fraction of the total applied dose when it removes a fraction of the amount still on the surface. The correction is to calculate \(fS(t_w^-)\), retain the rest on the surface, and leave the reservoir unchanged unless the model specifically represents wash-induced removal from skin. A second pitfall is to treat an altered rate constant as proof of a universal occlusion effect; the direction and size of the change depend on the formulation and conditions.

Check Your Understanding

Answer each question using the assumptions stated in it. Brief solutions follow the questions.

  1. For a no-loss model with \(S_0=20\) mg, \(k_p=0\), and \(R(0)=0\), what is eventually absorbed?
  2. A wash removes 25% of the surface amount at time \(t_w\). What happens immediately to \(S\), \(R\), and the recorded wash loss?
  3. If total dose is fixed and area doubles, must the eventual absorbed amount double under the no-loss model with positive \(k_p\) and \(k_r\)?
  4. Why can an input rate at a particular time not be substituted for the cumulative absorbed amount by that time?

Solutions. (1) Zero: with \(k_p=0\), no drug enters the reservoir, so \(S(t)=20\) mg remains on the surface and \(A(t)=0\). (2) \(S(t_w^+)=0.75S(t_w^-)\), \(R(t_w^+)=R(t_w^-)\) under the idealized wash assumption, and the removed amount \(0.25S(t_w^-)\) is recorded as surface loss. (3) No. With a fixed total dose, no surface loss, and positive transfer rates, the eventual absorbed amount is the fixed applied amount; area may change timing through \(k_p\), subject to the model assumptions. (4) The input rate \(k_rR(t)\) has units mg/h and is instantaneous, whereas cumulative absorption \(A(t)\) has units mg and is the time integral of that rate from the start of exposure.